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Page No 9:

Question 1:

What do you mean by Euclid's division algorithm.

Answer:

Euclid's division algorithm states that for any two positive integers a and b,there exist unique integers q and r, such that a = bq + r,where 0 ≤ r < b.

Page No 9:

Question 2:

A number when divided by 61 gives 27 as quotient and 32 as remainder.
Find the number.

(v) 30=2×3×572=2×2×2×3×3 = 23×32432=2×2×2×2×3×3×3 = 24×33
∴ HCF = Product of smallest power of each common prime factor in the numbers = 2×3=6
∴ ​LCM = Product of the greatest power of each prime factor involved in the numbers = 24×33×5=2160

(vi) 21=3×7
28 = 28=2×2×7 = 22×736=2×2×3×3 = 22×3245=5×3×3 = 5×32
∴​ HCF = Product of smallest power of each common prime factor in the numbers = 1
∴​ LCM = Product of the greatest power of each prime factor involved in the numbers = 22×32×5×7=1260

Page No 17:

Question 3:

The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other.

Answer:

Now, the greatest four digit number is 9999.
On dividing 9999 by 360 we get 279 as remainder.
Thus, 9999 − 279 = 9720 is exactly divisible by 360.

Hence, the greatest number of four digits which is exactly divisible by 15, 24 and 36 is 9720.

Page No 17:

Question 14:

In a seminar, the number of participants in Hindi, English and mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required, if in each room, the same number of participants are to be seated and all of them being in the same subject.

Answer:

HCF = product of smallest power of each common prime factor in the numbers = 22 × 3 = 12

Total number of paricipants = 60 + 84 + 108 = 252

Therefore, minimum number of rooms required = 25212=21

Thus, minimum number of rooms required is 21.

Page No 17:

Question 15:

Three sets of English, Mathematics and Science books containing 336, 240 and 96 books respectively have to be stacked in such a way that all the books are stored subject wise and the higher of each stack is the same. How many stacks will be there?

Answer:

Total number of English books = 336
Total number of mathematics books = 240
Total number of science books = 96
∴ Number of books stored in each stack = HCF (336, 240, 96)
Prime factorisation:
336 = 24×3×7
240 = 24×3×5
96 = 25×3
∴ HCF = Product of the smallest power of each common prime factor involved in the numbers = 24×3 = 48
Hence, we made stacks of 48 books each.

∴ Number of stacks = 33648+24048+9648 = (7 + 5 + 2) = 14

Page No 17:

Question 16:

Three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length. What is the greatest possible length of each plank?

Answer:

The lengths of three pieces of timber are 42 m, 49 m and 63 m, respectively.
We have to divide the timber into equal length of planks.
∴ Greatest possible length of each plank = HCF(42, 49, 63)
Prime factorisation:
42 = 2×3×7
49 = 7×7
63 = 3×3×7
∴ HCF = Product of smallest power of each common prime factor in the numbers = 7
Hence, the greatest possible length of each plank is 7 m.

Page No 17:

Question 17:

Find the greatest possible length which can be used to measure exactly the length 7 m, 3 m 85 cm and 12 m 95 cm.

Page No 18:

Question 18:

Find the maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils.

Answer:

Total number of pens = 1001
Total number of pencils = 910
∴​ Maximum number of students who get the same number of pens and pencils = HCF (1001, 910)
Prime factorisation:
1001 = 11×91
910 = 10×91
∴ HCF = 91
Hence, 91 students receive same number of pens and pencils.

Page No 18:

Question 19:

Find the leash number of square tiles required to pave the ceiling of a room 15 m 17 cm long and 9 m 2 cm broad.

Answer:

∴ ​LCM(2,4,6,8,10,12) = 23×3×5=120
Hence, after every 120 minutes (i.e. 2 hours), they will toll together.
∴ Required number of times = (302+1)=16

Page No 18:

Question 24:

Find the missing numbers in the following factorisation:

Answer:

660=2×2×3×5×11

Page No 25:

Question 1:

Without actual division, show that each of the following rational numbers is a terminating decimal. Express each in decimal form:

(i) 2323×52

(ii) 24125

(iii)171800

(iv) 151600

(v) 17320

(vi) 193125

Answer:

(i) 2323×52 = 23×523×53=1151000=0.115
We know either 2 or 5 is not a factor of 23, so it is in its simplest form.
Moreover, it is in the form of (2m×5n).
Hence, the given rational is terminating.

(ii) 24125=2453=24×2353×23=1921000=0.192
We know 5 is not a factor of 23, so it is in its simplest form.
Moreover, it is in the form of (2m×5n).
Hence, the given rational is terminating.

(iii) 171800 = 17125×52=171×5325×55=21375100000=0.21375
We know either 2 or 5 is not a factor of 171, so it is in its simplest form.
Moreover, it is in the form of (2m×5n).
Hence, the given rational is terminating.

(iv) 151600=1526×52=15×5426×56=93751000000=0.009375
We know either 2 or 5 is not a factor of 15, so it is in its simplest form.
Moreover, it is in the form of (2m×5n).
Hence, the given rational is terminating.

(v) 17320=1726×5 = 17×5526×56 = 531251000000=0.053125
We know either 2 or 5 is not a factor of 17, so it is in its simplest form.
Moreover, it is in the form of (2m×5n).
Hence, the given rational is terminating.

(vi) 193125 = 1955=19×2555×25=608100000=0.00608
We know either 2 or 5 is not a factor of 19, so it is in its simplest form.
Moreover, it is in the form of (2m×5n).
Hence, the given rational is terminating.

Page No 25:

Question 2:

Without actual division, show that each of the following rational numbers is a non-terminating repeating decimal:

Page No 35:

Question 1:

Answer:

Rational numbers: The numbers of the form pq where p,q are integers and q ≠ 0 are called rational numbers.
Example: 23
Irrational numbers: The numbers which when expressed in decimal form are expressible as non-terminating and non-repeating decimals are called irrational numbers.
Example: 2
Real numbers: The numbers which are positive or negative, whole numbers or decimal numbers and rational number or irrational number are called real numbers.
Example: 2, 13, 2, −3 etc.

Page No 35:

Question 3:

Answer:

(i) Let 6=2×3 be rational.
Hence, 2,3 are both rational.
This contradicts the fact that 2,3 are irrational.
The contradiction arises by assuming 6 is rational.
Hence, 6 is irrational.

(ii) Let 2-3 be rational.
Hence, 2 and 2-3 are rational.
∴ (2-2+3)=3 = rational [∵ Difference of two rational is rational]
This contradicts the fact that 3 is irrational.
The contradiction arises by assuming 2-3 is rational.
Hence, 2-3 is irrational.

(iii) Let 3+2 be rational.
Hence, 3 and 3+2 are rational.
∴ 3+2-3=2 = rational [∵ Difference of two rational is rational]
This contradicts the fact that 2 is irrational.
The contradiction arises by assuming 3+2 is rational.
Hence, 3+2 is irrational.

(iv) Let 2+5 be rational.
Hence, 2+5 and 5 are rational.
∴ (2+5)-2=2+5-2=5 = rational [∵ Difference of two rational is rational]
This contradicts the fact that 5 is irrational.
The contradiction arises by assuming 2-5 is rational.
Hence, 2-5 is irrational.

(v) Let, 5+32 be rational.
Hence, 5 and 5+32 are rational.
∴ (5+32-5)=32 = rational [∵ Difference of two rational is rational]
∴ 13×32=2 = rational [∵ Product of two rational is rational]
This contradicts the fact that 2 is irrational.
The contradiction arises by assuming 5+32 is rational.
Hence, 5+32 is irrational.

(vi) Let 37 be rational.
∴ 13×37=7 = rational [∵ Product of two rational is rational]
This contradicts the fact that 7 is irrational.
The contradiction arises by assuming 37 is rational.
Hence, 37 is irrational.

(vii) Let 35 be rational.
∴ 13×35=15 = rational [∵ Product of two rational is rational]
This contradicts the fact that 15 is irrational.
∴ 1×55×5=155
So, if 15 is rational, then 155 is rational.
∴ 5×155=5 = rational [∵ Product of two rational is rational]
Hence, 15 is irrational.
The contradiction arises by assuming 35 is rational.
Hence, 35 is irrational.

(viii) Let 2-35 be rational.
Hence 2 and 2-35 are rational.
∴ 2-(2-35)=2-2+35=35 = rational [∵ Difference of two rational is rational]
∴ 13×35=5 = rational [∵ Product of two rational is rational]
This contradicts the fact that 5 is irrational.
The contradiction arises by assuming 2-35 is rational.
Hence, 2-35 is irrational.

(ix) Let 3+5 be rational.
∴ 3+5=a, where a is rational
∴ 3=a-5 ... (1)
On squaring both sides of equation (1), we get3=(a-5)2=a2+5-25a
⇒ 5=a2+22a
This is impossible because right-hand side is rational, whereas the left-hand side is irrational.
This is a contradiction.
Hence, 3+5 is irrational.

Page No 35:

Question 4:

Prove that 13 is irrational.

Answer:

Let 13 be rational.
∴ 13 = ab, where a,b are positive integers having no common factor other than 1
∴ 3=ba ...(1)
Since a,b are non-zero integers, ba is rational.
Thus, equation (1) shows that 3 is rational.
This contradicts the fact that 3 is rational.
The contradiction arises by assuming 3 is rational.
Hence, 13 is irrational.

Page No 35:

Question 5:

(i) Give an example of two irrationals whose sum is rational.
(ii) Give an examples of two irrationals whose product is rational.

Answer:

Page No 35:

Question 6:

State whether the given statement is true of false:

(i) The sum of two rationals is always rational.
(ii) The product of two rationals is always rational.
(iii) The sum of two irrationals is an irrational.
(iv) The product of two irrationals is an irrational.
(v) The sum of a rational and and irrational is irrational.
(vi) The product of a rational and an irrational is irrational.

Page No 36:

Question 7:

Prove that (23-1) is an irrational number.

Answer:

Let x = 23-1 be a rational number.

x=23-1⇒x2=23-12⇒x2=232+12-2231⇒x2=12+1-43⇒x2-13=-43⇒13-x24=3
Since x is a rational number, x2 is also a rational number.
⇒ 13 − x2 is a rational number
⇒ 13-x24 is a rational number
⇒ 3 is a rational number
But 3 is an irrational number, which is a contradiction.
Hence, our assumption is wrong.

Thus, (23-1) is an irrational number.

Page No 36:

Question 8:

Prove that (4-52) is an irrational number.

Answer:

​Let x = 4-52 be a rational number.

x=4-52⇒x2=4-522⇒x2=42+522-2452⇒x2=16+50-402⇒x2-66=-402⇒66-x240=2
Since x is a rational number, x2 is also a rational number.
⇒ 66 − x2 is a rational number
⇒ 66-x240 is a rational number
⇒ 2 is a rational number
But 2 is an irrational number, which is a contradiction.
Hence, our assumption is wrong.

Thus, (4-52) is an irrational number.

Page No 36:

Question 9:

Prove that (5-23) is an irrational number.

Answer:

​​Let x = 5-23 be a rational number.

x=5-23⇒x2=5-232⇒x2=52+232-2523⇒x2=25+12-203⇒x2-37=-203⇒37-x220=3
Since x is a rational number, x2 is also a rational number.
⇒ 37 − x2 is a rational number
⇒ 37-x220 is a rational number
⇒ 3 is a rational number
But 3 is an irrational number, which is a contradiction.
Hence, our assumption is wrong.

Thus, (5-23) is an irrational number.

Page No 36:

Question 10:

Prove that 52 is irrational.

Answer:

Let 52 is a rational number.

∴ 52=pq, where p and q are some integers and HCF(p, q) = 1 ....(1)⇒52q=p⇒52q2=p2⇒225q2=p2
⇒ p2 is divisible by 2
⇒ p is divisible by 2 .....(2)

Page No 36:

Question 2:

Answer:

The fundamental theorem of arithmetic, states that every integer greater than 1 either is prime itself or is the product of prime numbers, and this product is unique.

Page No 36:

Question 3:

Express 360 as product of its prime factors.

Answer:

Prime factorization:

360 = 23 × 32 × 5

Page No 36:

Question 4:

If a and b are two prime numbers then find HCF(a, b).

Answer:

Prime factorization:a = ab = b

HCF = product of smallest power of each common prime factor in the numbers = 1

Thus, HCF(a, b) = 1

Page No 36:

Question 5:

If a and b are two prime numbers then find LCM(a, b).

Answer:

Prime factorization:a = ab = b

LCM = product of greatest power of each prime factor involved in the numbers = a × b

Thus, LCM(a, b) = ab.

Page No 36:

Question 6:

If the product of two numbers is 1050 and their HCF is 25, find their LCM.

Answer:

HCF of two numbers = 25
Product of two numbers = 1050
Let their LCM be x.

Using the formula, Product of two numbers = HCF × LCM
we conclude that,

1050 = 25 × xx = 105025
= 42

Hence, their LCM is 42.

Page No 36:

Question 7:

What is a composite number?

Answer:

A composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself).

Page No 36:

Question 8:

If a and b are relatively prime then what is their HCF?

Answer:

If two numbers are relatively prime then their greatest common factor will be 1.

Thus, HCF(a, b) = 1.

Page No 36:

Question 9:

If the rational number ab has a terminating decimal expansion, what is the condition to be satisfied by b?

Answer:

Let x be a rational number whose decimal expansion terminates.
Then, we can express x in the form ab, where a and b are coprime, and prime factorization of b is of the form (2m × 5n), where m and n are non negative integers.

Page No 36:

Question 10:

Simplify: 245+32025.

Answer:

245+32025=23×3×5+32×2×525=2×35+3×2525=65+6525=12525=6

Thus, simplified form of 245+32025 is 6.

Page No 36:

Question 11:

Write the decimal expansion of 7324×53.

Answer:

Decimal expansion:7324×53=73×524×54=3652×54=365104=36510000=0.0365

Thus, the decimal expansion of 7324×53 is 0.0365.

Page No 37:

Question 12:

Show that there is no value of n for which (2n × 5n) ends in 5.

Answer:

We can write:
(2n × 5n) = (2 × 5)n
= 10n

For any value of n, we get 0 in the end.

Thus, there is no value of n for which (2n × 5n) ends in 5.

Page No 37:

Question 13:

Is it possible to have two numbers whose HCF is 25 and LCM is 520?

Answer:

​No, it is not possible to have two numbers whose HCF is 25 and LCM is 520.

Since, HCF must be a factor of LCM, but 25 is not a factor of 520.

Page No 37:

Question 14:

Give an example of two irrationals whose sum is rational.

Answer:

Let the two irrationals be 4-5and4+5.

4-5+4+5=8

​Thus, sum (i.e., 8) is a rational number.

Page No 37:

Question 15:

Give an example of two irrationals whose product is rational.

Answer:

​​Let the two irrationals be 45and35.

45×35=60

​Thus, product (i.e., 60) is a rational number.

Page No 37:

Question 16:

If a and b are relatively prime, what is their LCM?

Answer:

If two numbers are relatively prime then their greatest common factor will be 1.

∴ HCF(a, b) = 1

Using the formula, Product of two numbers = HCF × LCM
we conclude that,

a × b = 1 × LCM
∴ LCM = ab

Thus, LCM(a, b) is ab.

Page No 37:

Question 17:

The LCM of two numbers is 1200. Show that the HCF of these numbers cannot be 500. Why?

Answer:

If the LCM of two numbers is 1200 then, it is not possible to have their HCF equals to 500.

Since, HCF must be a factor of LCM, but 500 is not a factor of 1200.

Page No 37:

Question 18:

Express 0.4¯ as a rational number in simplest form.

Answer:

Let x be 0.4¯.

x=0.4¯ .....(1)
Multiplying both sides by 10, we get10x=4.4¯ .....(2)

Subtracting (1) from (2), we get10x-x=4.4¯-0.4¯⇒9x=4⇒x=49

Thus, simplest form of 0.4¯ as a rational number is 49.

Page No 37:

Question 19:

Express 0.23¯ as a rational number in simplest form.

Answer:

​Let x be 0.23¯.

x=0.23¯ .....(1)
Multiplying both sides by 100, we get100x=23.23¯ .....(2)

From (2) and (3), 2 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.

Thus, 23 is irrational.

Page No 37:

Question 22:

Write a rational number between 3and2.

Answer:

Since, 3 = 1.732....
So, we may take 1.8 as the required rational number between 3and2.

Thus, the required rational number is 1.8

Page No 37:

Question 23:

Explain why 3.1416¯ is a rational number.

Answer:

Since, 3.1416¯ is a non-terminating repeating decimal.

Hence, is a rational number.

Page No 38:

Question 1:

Which of the following rational numbers is expressible as a now terminating repeating decimal?

(a) 13511250
(b) 2017250
(c) 32191800
(d) 1723625

Answer:

(c) 32191800

13511250=135154×2
We know 2 and 5 are not the factors of 1351.
So, the given rational is in its simplest form.
And it is of the form (2m×5n) for some integers m,n.
So, the given number is a terminating decimal.
∴ 135154×2=1351×254×24=1080810000=1.0808

2017250 = 201753×2
We know 2 and 5 are not the factors of 2017.
So, the given rational number is in its simplest form.
And it is of the form (2m×5n) for some integers m,n.
So, the given rational number is a terminating decimal.
∴ 201753×2=2017×2253×23=80681000=8.068

32191800=321923×52×32
We know 2, 3 and 5 are not the factors of 3219.
So, the given rational number is in its simplest form.
∴ (23×52×32) ≠ (2m×5n)
Hence, 32191800 is not a terminating decimal.32191800=1.78833333.....
Thus, it is a repeating decimal.

1723625=172354
We know 5 is not a factor of 1723.
So, the given rational number is in its simplest form.
And it is not of the form (2m×5n).
Hence, 1723650 is not a terminating decimal.

Page No 38:

Question 2:

If a = (22 × 33 × 54) and b = (23 × 32 × 5), then HCF (a, b) = ?

(a) 90
(b) 180
(c) 360
(d) 540

Answer:

(b) 180
It is given that:a = (22×33×54) and b = (23×32×5)
∴ HCF (a, b) = Product of smallest power of each common prime factor in the numbers
= 22×32×5
= 180

Page No 38:

Question 3:

HCF of (23 × 32 × 5), (22 × 33 ×52) and (24 ×3 × 53 × 7) is

(a) 30
(b) 48
(c) 60
(d) 105

Answer:

(c) 60

HCF = (23×32×5, 22×33×52, 24×3×53×7)
HCF = Product of smallest power of each common prime factor in the numbers
= 22×3×5
= 60

Page No 38:

Question 10:

Answer:

Page No 38:

Question 11:

Euclid's division lemma sates that for any positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy

(a) 1 < r < b
(b) 0 < r ≤ b
(c) 0 ≤ r < b
(d) 0 < r < b

Answer:

(c) 0 ≤ r < b

Euclid's division lemma states that for any positive integers a and b,there exist unique integers q and r such that a = bq + r,
where r​ must satisfy 0 ≤ r < b

Page No 39:

Question 12:

A number when divided by 143 leaves 31 as remainder. What will be the remainder when the same number is divided by 13?

(a) 0
(b) 1
(c) 3
(d) 5

Answer:

(d) 5

We know,
Dividend = Divisor × Quotient + Remainder.
It is given that:
Divisor = 143
Remainder = 13
So, the given number is in the form of 143x + 31, where x is the quotient.
∴ 143x + 31 = 13 (11x) + (13 ×2) + 5 = 13 (11x+ 2) + 5
Thus, the remainder will be 5 when the same number is divided by 13.

Page No 39:

Question 13:

Which of the following is an irrational number?

(a) 227
(b) 3.1416
(c) 3.1416
(d) 3.141141114...

Answer:

(d) 3.141141114...

3.141141114 is an irrational number because it is a non-repeating and non-terminating decimal.

Page No 39:

Question 14:

π is

(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these

Answer:

(c) an irrational number

π is an irrational number because it is a non-repeating and non-terminating decimal.

Page No 39:

Question 15:

2.35 is
(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these

Answer:

(b) a rational number
2.35 is a rational number because it is a repeating decimal.

Page No 39:

Question 16:

2.13113111311113...is

(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these

Answer:

(c) an irrational number

It is an irrational number because it is a non-terminating and non-repeating decimal.

Page No 39:

Question 17:

3.24636363...is

(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these

Answer:

(b) a rational number

It is a rational number because it is a repeating decimal.

Page No 39:

Question 18:

Which of the following rational numbers is expressible as a terminating decimal?

(a) 124165
(b) 13130
(c) 2027625
(d) 1625462

Answer:

(c) 2027625

124165=1245×33; we know 5 and 33 are not the factors of 124. It is in its simplest form and it cannot be expressed as the product of (2m×5n) for some non-negative integers m,n.

So, it cannot be expressed as a terminating decimal.

13130 = 1315×6; we know 5 and 6 are not the factors of 131. Its is in its simplest form and it cannot be expressed as the product of ( 2m×5n) for some non-negative integers m,n.

So, it cannot be expressed as a terminating decimal.

2027625=2027×2454×24=3243210000=3.2432; as it is of the form (2m×5n), where m,n are non-negative integers.
So, it is a terminating decimal.

1625462=16252×7×33; we know 2, 7 and 33 are not the factors of 1625. It is in its simplest form and cannot be expressed as the product of (2m×5n) for some non-negative integers m,n.
So, it cannot be expressed as a terminating decimal.

Page No 39:

Question 19:

The decimal expansion of the rational number 3722×5 will terminate after

(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal places

Answer:

(b) two decimal places

3722×5=37×522×52=185100=1.85

So, the decimal expansion of the rational number will terminate after two decimal places.

Page No 39:

Question 20:

The decimal expansion of the number 147531250 will terminate after

(a) one decimal place
(b) two decimal place
(c) three decimal place
(d) four decimal place

Answer:

(d) four decimal places

147531250=1475354×2=14753×2354×24=11802410000 = 11.8024

So, the decimal expansion of the number will terminate after four decimal places.

Page No 39:

Question 21:

The number 1.732 is
(a) an irrational number
(b) a rational number
(c) an integer
(d) a whole number

Answer:

​Clearly, 1.732 is a terminating decimal.

Hence, a rational number.

Hence, the correct answer is option (b).

Page No 40:

Question 22:

a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5. Then, the least prime factor of (a + b) is

(a) 2
(b) 3
(c) 5
(d) 8

Answer:

(a) 2

Since 5 + 3 = 8, the least prime factor of a + b has to be 2, unless a + b is a prime number greater than 2.
If a + b is a prime number greater than 2, then a + b must be an odd number. So, either a or b must be an even number. If a is even, then the least prime factor of a is 2, which is not 3 or 5. So, neither a nor b can be an even number. Hence, a + b cannot be a prime number greater than 2 if the least prime factor of a is 3 or 5.

Page No 40:

Question 23:

2 is
(a) a rational number
(b) an irrational number
(c) a terminating decimal
(d) a nonterminating repeating decimal

Answer:

Let 2 is a rational number.

∴ 2=pq, where p and q are some integers and HCF(p, q) = 1 .... (1)⇒2q=p⇒2q2=p2⇒2q2=p2
⇒ p2 is divisible by 2
⇒ p is divisible by 2 .... (2)

From (2) and (3), 2 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.

Thus, 2 is an irrational number.

Hence, the correct answer is option (b).

Page No 40:

Question 24:

12is

(a) a fraction
(b) a rational number
(c) an irrational number
(d) none of these

Answer:

(c) an irrational number

12 is an irrational number.

Page No 40:

Question 25:

2+2 is
(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these

Answer:

(c) an irrational number

2+2 is an irrational number.
if it is rational, then the difference of two rational is rational
∴​ (2+2)-2 = 2 = irrational

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Question 26:

What is the least number that divisible by all the natural numbers from 1 to 10 (both inclusive)?

(a) 100
(b) 1260
(c) 2520
(d) 5040

Answer:

(c) 2520

We have to find the least number that is divisible by all numbers from 1 to 10.
∴ LCM (1 to 10) = 23×32×5×7=2520
Thus, 2520 is the least number that is divisible by every element and is equal to the least common multiple.

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Question 1:

The decimal representation of 71150is
(a) a terminating decimal
(b) a non-terminating, repeating decimal
(b) a non-terminating and non-repeating decimal
(d) none of these

Answer:

(b) a non-terminating, repeating decimal

71150=712×3×52
We know that 2, 3 or 5 are not factors of 71.
So, it is in its simplest form.
And, (2×3×52) ≠ (2m×5n)
∴ 71150=0.473¯
Hence, it is a non-terminating, repeating decimal.

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Question 2:

Which of the following has terminating decimal expansion?

(a) 3291
(b) 1980
(c) 2345
(d) 2542

Answer:

(b) 1980

1980=1924×5
We know 2 and 5 are not factors of 19, so it is in its simplest form.
And (24×5)=(2m×5n)
Hence, 1980 is a terminating decimal.

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Question 3:

On dividing a positive integer n by 9, we get 7 as remainder. What will be the remainder if (3n − 1) is divided by 9?

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Question 4:

0.68+0.73=?
(a) 1.41
(b) 1.42
(c) 0.141
(d) None of these

Answer:

(b) 1.42

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Question 5:

Show that any number of the form 4n, n ∈ N can never end with the digit 0.

Answer:

If 4n ends with 0, then it must have 5 as a factor.
But we know the only prime factor of 4n is 2.
Also we know from the fundamental theorem of arithmetic that prime factorisation of each number is unique.
Hence, 4n can never end with the digit 0.

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Question 6:

The HCF of two numbers is 27 and their LCM is 162. If one of the number is 81, find the other.

Answer:

(c) 3.142857 is rational because it is a terminating decimal.
(d) 2.3 is rational because it is a non-terminating, repeating decimal.

(e) π is irrational because it is a non-repeating, non-terminating decimal.

(f) 227 is rational because it is in the form of pq,q ≠ 0.

(g) 0.232332333... is irrational because it is a non-terminating, non-repeating decimal.
(h) 5.2741 is rational because it is a non-terminating, repeating decimal.

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Question 10:

Prove that 2+3 is irrational.

Answer:

Let (2+3) be rational.
Then, both (2+3) and 2 are rational.
∴ {(2+3)-2} is rational [∵ Difference of two rational is rational]
⇒ 3 is rational.
This contradicts the fact that 3 is irrational.
The contradiction arises by assuming (2+3) is rational.
Hence, (2+3) is irrational.